Question

Super Sneaker Company is evaluating two different materials, A and B, to be used to construct the soles of their new active shoe targeted to city high school students in Canada. While material B costs less than material A, the company suspects that mean wear for material B is greater than mean wear for material A. Two study designs were initially developed to test this suspicion. In both designs, Halifax was chosen as a representative city of the targeted market. In Study Design 1, 14 high school students were drawn at random from the Halifax School District database. After obtaining their shoe sizes, the company manufactured 14 pairs of shoes, each pair with one shoe having a sole constructed from material A and the other shoe, a sole constructed from material B.

Answer 1

The test will be the t-test for matched pairs.

The null hypothesis is: H0:μB−μA=0

b.

The alternative hypothesis is: H1:μB−μA>0

c.

The test statistic is - 6.6764.

First, calculate the mean difference and the standard deviation of the difference in scores. The table at the bottom of the page shows the summary of those calculations. The Mean difference, ¯D is sum of the differences divided by the sample size. That is:

¯D=3.879=0.43

The equation for the standard deviation of the differences, SD is:

SD=(Σ(D−¯D)2n−1).5

Where:

• SD is the sample standard deviation of differences

• D is the value of one of the samples

• ¯D is the sample mean difference

• n is the sample size

Substituting values from the table:

SD=(24.19188).5 = 1.7390

The equation for the test statistic is:

tstat=D¯−μDSD/n

Substituting values:

tstat=3.871.7390/9 = 6.6764

d.

The degrees of freedom are 8.

The degrees of freedom is the sample size minus 1. n−1=9−1=8

e.

This is a one sided test because the alternative hypothesis is "greater than".

f.

The 90% confidence interval is: μdiff=3.87±1.0779 or [2.7921,4.9479]. The value of zero is not in the confidence interval, therefore there is sufficient evidence to conclude that the Material B is performing better than Material A.

The t table is set up for a right tailed probability, and the confidence interval is a two tailed probability. Thus, to find the t value go to the row for n - 1 = 8 degrees of freedom and over to the column for (1+0.90)2=95% confidence. That will be correct for a 90% confidence interval.

So, the t statistic for 90% confidence and 8 degrees of freedom is t=1.0779

The equation for the confidence interval is: μdiff=μB−μA±t∗s√n

Substituting values:

μdiff=3.87±1.8595∗1.7390√9μdiff=3.87±1.0779 or [2.7921,4.9479]

Student	Material A	Material B	D	D-Dbar	{D- Dbar)^2
1	18.56	16.67	1.89	2.3200	5.3824
2	15.39	17.01	-1.62	-1.1900	1.4161
3	13.9	15.91	-2.01	-1.5800	2.4964
4	16.87	15.41	1.46	1.8900	3.5721
5	14.57	17.27	-2.7	-2.2700	5.1529
6	14.31	16.19	-1.88	-1.4500	2.1025
7	15.8	15.13	0.67	1.1000	1.2100
8	15.67	16.55	-0.88	-0.4500	0.2025
9	17.72	16.52	1.2	1.6300	2.6569
Sum			-3.87		24.1918
Mean, S.D.			-0.4300		1.7390